Rationalizing Denominators in Rational Expressions

Fully rationalize and simplify the following fractions:

1. $\dfrac{4}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{4\sqrt{6}}{(\sqrt{6})(\sqrt{6})}=\dfrac{4\sqrt{6}}{6}=\dfrac{2\sqrt{6}}{3}$

2. $\dfrac{x}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{x\sqrt{3}}{(\sqrt{3})(\sqrt{3})}=\dfrac{x\sqrt{3}}{3}$

3. $\dfrac{2}{\sqrt{12}} \times \dfrac{\sqrt{12}}{\sqrt{12}}=\dfrac{2\sqrt{12}}{(\sqrt{12})(\sqrt{12})}=\dfrac{2\sqrt{12}}{12}=\dfrac{4\sqrt{3}}{12}=\dfrac{\sqrt{3}}{3}$

4. $\dfrac{3}{i} \times \dfrac{i}{i}=\dfrac{3i}{i \times i}=\dfrac{3i}{i^2}=\dfrac{3i}{-1}=-3i$

5. $\dfrac{2}{6i} \times \dfrac{i}{i}=\dfrac{2i}{6i \times i}=\dfrac{2i}{6i^2}=\dfrac{2i}{-6}=-\dfrac{i}{3}$

6. $\dfrac{5}{\sqrt{6}+2} \times \dfrac{\sqrt{6}-2}{\sqrt{6}-2}=\dfrac{5(\sqrt{6}-2)}{(\sqrt{6}+2)(\sqrt{6}-2)}=\dfrac{5\sqrt{6}-10}{\sqrt{6}^2-4}=\dfrac{5\sqrt{6}-10}{6-4}=\dfrac{5\sqrt{6}-10}{2}$

7. $\dfrac{1}{\sqrt{3}-5} \times \dfrac{\sqrt{3}+5}{\sqrt{3}+5}=\dfrac{\sqrt{3}+5}{(\sqrt{3}-5)(\sqrt{3}+5)}=\dfrac{\sqrt{3}+5}{\sqrt{3}^2-25}=\dfrac{\sqrt{3}+5}{3-25}=\dfrac{\sqrt{3}+5}{-22}$

8. $\dfrac{1}{4+2\sqrt{4}}=\dfrac{1}{4+2(2)}=\dfrac{1}{8}$

Note: If you can find the square root in the denominator, you don't have to rationalize! Eliminate the radical by finding the square root.

9. $\dfrac{1}{5-3\sqrt{2}} \times \dfrac{5+3\sqrt{2}}{5+3\sqrt{2}}=\dfrac{5+3\sqrt{2}}{(5-3\sqrt{2})(5+3\sqrt{2})}=\dfrac{5+3\sqrt{2}}{25-9(\sqrt{2}^2)}=\dfrac{5+3\sqrt{2}}{25-18}=\dfrac{5+3\sqrt{2}}{7}$

10. $\dfrac{3}{i+\sqrt{7}}  \times \dfrac{i-\sqrt{7}}{i-\sqrt{7}}=\dfrac{3(i-\sqrt{7}}{(i+\sqrt{7})(i-\sqrt{7})}=\dfrac{3i-3\sqrt{7}}{i^2-\sqrt{7}^2}=\dfrac{3i-3\sqrt{7}}{-1-7}=\dfrac{3i-3\sqrt{7}}{-8}$